Answer
The $x$ and $y$ components of the resultant force are zero. Thus the resultant force is zero (see in step by step solution).
Work Step by Step
First, we resolve each force into its $x$ and $y$ components by using the parallelogram law:
Let $F_{1x}$ and $F_{1y}$ are the $x$ and $y$ components of $\vec {F_1}$ force, in which $F_1=8\;kN$. Using proportional parts of similar triangles and considering the directions, we have
$\frac{F_{1x}}{8\;kN}=\frac{4}{5}$
or, $F_{1x}=8\;kN\Big(\frac{4}{5}\Big)=6.4\;kN$
Similarly,
$\frac{F_{1y}}{8\;kN}=-\frac{3}{5}$
or, $F_{1y}=-8\;kN\Big(\frac{3}{5}\Big)=-4.8\;kN$
Let $F_{2x}$ and $F_{2y}$ are the $x$ and $y$ components of $\vec {F_2}$ force, in which $F_2=6\;kN$. Using proportional parts of similar triangles and considering the directions, we have
$\frac{F_{2x}}{6\;kN}=\frac{3}{5}$
or, $F_{2x}=6\;kN\Big(\frac{3}{5}\Big)=3.6\;kN$
Similarly,
$\frac{F_{2y}}{6\;kN}=\frac{4}{5}$
or, $F_{2y}=6\;kN\Big(\frac{4}{5}\Big)=4.8\;kN$
Let $F_{3x}$ and $F_{3y}$ are the $x$ and $y$ components of $\vec {F_3}$ force, in which $F_3=4\;kN$. Thus we have
$F_{3x}=-4\cos0^\circ\;kN=-4\;kN$
$F_{3y}=4\sin0^\circ\;kN=0\;kN$
Let $F_{4x}$ and $F_{4y}$ are the $x$ and $y$ components of $\vec {F_4}$ force, in which $F_4=6\;kN$. Thus we have
$F_{4x}=-6\cos0^\circ\;kN=-6\;kN$
$F_{4y}=6\sin0^\circ\;kN=0\;kN$
Second, using scalar notation and indicating the positive directions of components along the $x$ and $y$ axes with symbolic arrows, we sum these components algebraically:
Summing the x components, we have
$\xrightarrow{+} (F_R)_x=\sum F_x$
or, $(F_R)_x=(6.4+3.6-4-6)\;kN=0\;kN$
Summing the y components yields
$+\uparrow (F_R)_y=\sum F_y$
or, $(F_R)_y=(-4.8+4.8+0+0)\;kN=0\;kN$
$\therefore$ The $x$ and $y$ components of the resultant force are zero. Thus the resultant force is zero.
